In figure, O is the centre of the circle, BO is the bisector of ∠ABC

1.	In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.
Answer» Since, BO is the bisector of ∠ABC, then, ∠ABO = ∠CBO …..(i) From figure: Radius of circle = OB = OA = OB = OC ∠OAB = ∠OCB …..(ii) [opposite angles to equal sides] ∠ABO = ∠DAB …..(iii) [opposite angles to equal sides] From equations (i), (ii) and (iii), we get ∠OAB = ∠OCB …..(iv) In ΔOAB and ΔOCB: ∠OAB = ∠OCB [From (iv)] OB = OB [Common] ∠OBA = ∠OBC [Given] Then, By AAS condition : ΔOAB ≅ ΔOCB So, AB = BC [By CPCT]

In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.

Answer»

Since, BO is the bisector of ∠ABC, then,

∠ABO = ∠CBO …..(i)

From figure:

Radius of circle = OB = OA = OB = OC

∠OAB = ∠OCB …..(ii) [opposite angles to equal sides]

∠ABO = ∠DAB …..(iii) [opposite angles to equal sides]

From equations (i), (ii) and (iii), we get

∠OAB = ∠OCB …..(iv)

In ΔOAB and ΔOCB:

∠OAB = ∠OCB [From (iv)]

OB = OB [Common]

∠OBA = ∠OBC [Given]

Then, By AAS condition :

ΔOAB ≅ ΔOCB

So, AB = BC [By CPCT]

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In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.

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